## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 10} (3-\frac{4}{5}x) = -5$
Let $\epsilon \gt 0$ be given. Let $\delta = \frac{5\epsilon}{4}$ Suppose that $\vert x-10 \vert \lt \delta$. Then: $\vert (3-\frac{4}{5}x)-(-5) \vert = \vert 8-\frac{4}{5}x \vert = \vert \frac{40-4x}{5} \vert = \frac{4}{5}\vert 10-x \vert \lt \frac{4}{5}\delta = \frac{4}{5}(\frac{5\epsilon}{4}) = \epsilon$ Therefore, $\lim\limits_{x \to 10} (3-\frac{4}{5}x) = -5$